The convergence of an implicit mean value iteration

Stefan Soltuz
(original), paper

Abstract

We show the convergence of an implicit mean value iteration when applied to uniformly pseudocontractive maps. Remarks about other implicit mean value iterations are given

Authors

Stefan M. Soltuz
Tiberiu Popoviciu Institute of Numerical Analysis

B. E. Rhoades
Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA

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B.E. Rhoades, S.M. Soltuz, The convergence of an implicit mean value iteration,  International Journal of Mathematics and Mathematical Sciences, 2006 ID 68369, 7 p.

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2006 Soltuz MC – The convergence of an implicit mean value iteration

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International Journal of Mathematics and Mathematical Sciences

 

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Wiley

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0161-1712

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1687-0425

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[8] S¸ . M. Soltuz, The backward Mann iteration, Octogon Mathematics Magazine 9 (2001), no. 2,797–800.
[9] , New technique for proving the equivalence of Mann and Ishikawa iterations, Revued’Analyse Numerique et de Theorie de l’Approximation 34 (2005), no. 1, 103–108.
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2006-Soltuz-MC-The-convergence-of-an-implicit-mean-value-iteration-

THE CONVERGENCE OF AN IMPLICIT MEAN VALUE ITERATION

B. E. RHOADES AND ŞTEFAN M. ŞOLTUZ

Received 3 April 2006; Revised 10 May 2006; Accepted 11 May 2006
We show the convergence of an implicit mean value iteration when applied to uniformly pseudocontractive maps. Remarks about other implicit mean value iterations are given.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.

1. Introduction

Let X X XXX be a real Banach space, T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a map, and x 0 , u 0 ∈ X x 0 , u 0 ∈ X x_(0),u_(0)in Xx_{0}, u_{0} \in Xx0,u0∈X. In [5], the following iteration is introduced:
(1.1) u n + 1 = ( 1 − α n ) u n + α n T u n , (1.1) u n + 1 = 1 − α n u n + α n T u n , {:(1.1)u_(n+1)=(1-alpha_(n))u_(n)+alpha_(n)Tu_(n)",":}\begin{equation*} u_{n+1}=\left(1-\alpha_{n}\right) u_{n}+\alpha_{n} T u_{n}, \tag{1.1} \end{equation*}(1.1)un+1=(1−αn)un+αnTun,
where { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1).
In the rest of the paper, we will assume that ( I − t T ) − 1 ( I − t T ) − 1 (I-tT)^(-1)(I-t T)^{-1}(I−tT)−1 exists for all t ∈ ( 0 , 1 ) t ∈ ( 0 , 1 ) t in(0,1)t \in(0,1)t∈(0,1). Consider the following iteration, see [8]:
(1.2) x n + 1 = ( 1 − α n ) x n + α n T x n + 1 , (1.2) x n + 1 = 1 − α n x n + α n T x n + 1 , {:(1.2)x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Tx_(n+1)",":}\begin{equation*} x_{n+1}=\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T x_{n+1}, \tag{1.2} \end{equation*}(1.2)xn+1=(1−αn)xn+αnTxn+1,
where the sequence { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} is in ( 0,1 ).
The following example shows that iteration (1.2) is well defined. We recall the following well-known result.
Lemma 1.1. Let { β n } β n {beta_(n)}\left\{\beta_{n}\right\}{βn} be a nonnegative sequence such that β n ∈ ( 0 , 1 ] β n ∈ ( 0 , 1 ] beta_(n)in(0,1]\beta_{n} \in(0,1]βn∈(0,1], for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N. If ∑ n = 1 ∞ β n = ∞ ∑ n = 1 ∞   β n = ∞ sum_(n=1)^(oo)beta_(n)=oo\sum_{n=1}^{\infty} \beta_{n}=\infty∑n=1∞βn=∞, then ∏ n = 1 ∞ ( 1 − β n ) = 0 ∏ n = 1 ∞   1 − β n = 0 prod_(n=1)^(oo)(1-beta_(n))=0\prod_{n=1}^{\infty}\left(1-\beta_{n}\right)=0∏n=1∞(1−βn)=0.
Inspired by [3, 8], we give an example which shows that Mann iterations (1.1) and (1.2) are independent.
Example 1.2. Let X = R 2 X = R 2 X=R^(2)X=\mathbb{R}^{2}X=R2. Let T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X be the map given by
(1.3) T ( x , y ) = ( 1 1 − 1 1 ) ( x y ) , ∀ ( x , y ) ∈ R 2 . (1.3) T ( x , y ) = 1 1 − 1 1 ( x y ) , ∀ ( x , y ) ∈ R 2 . {:(1.3)T(x","y)=([1,1],[-1,1])((x)/(y))","quad AA(x","y)inR^(2).:}T(x, y)=\left(\begin{array}{cc} 1 & 1 \tag{1.3}\\ -1 & 1 \end{array}\right)\binom{x}{y}, \quad \forall(x, y) \in \mathbb{R}^{2} .(1.3)T(x,y)=(11−11)(xy),∀(x,y)∈R2.
Iteration (1.1) is not convergent to the fixed point of T T TTT, while iteration (1.2), for { α n } ⊂ ( 1 / 2 , 1 ) α n ⊂ ( 1 / 2 , 1 ) {alpha_(n)}sub(1//2,1)\left\{\alpha_{n}\right\} \subset (1 / 2,1){αn}⊂(1/2,1), converges to the fixed point of T T TTT.
Proof. Let u = ( x , y ) u = ( x , y ) u=(x,y)u=(x, y)u=(x,y). For all λ ∈ ] 0 , 1 [ λ ∈ ] 0 , 1 [ lambda in]0,1[\lambda \in] 0,1[λ∈]0,1[,
| ( 1 − λ ) u + λ T u | 2 = | ( 1 − λ ) ( x , y ) + λ T ( x , y ) | 2 = | ( 1 − λ ) ( x , y ) + λ ( x + y , − x + y ) | 2 (1.4) = | ( 1 − λ ) x + λ x + λ y , ( 1 − λ ) y − λ x + λ y | 2 = | ( x + λ y , y − λ x ) | 2 = ( 1 + λ 2 ) x 2 + ( 1 + λ 2 ) y 2 = ( 1 + λ 2 ) | u | 2 > | u | 2 . | ( 1 − λ ) u + λ T u | 2 = | ( 1 − λ ) ( x , y ) + λ T ( x , y ) | 2 = | ( 1 − λ ) ( x , y ) + λ ( x + y , − x + y ) | 2 (1.4) = | ( 1 − λ ) x + λ x + λ y , ( 1 − λ ) y − λ x + λ y | 2 = | ( x + λ y , y − λ x ) | 2 = 1 + λ 2 x 2 + 1 + λ 2 y 2 = 1 + λ 2 | u | 2 > | u | 2 . {:[|(1-lambda)u+lambda Tu|^(2)=|(1-lambda)(x","y)+lambda T(x","y)|^(2)],[=|(1-lambda)(x","y)+lambda(x+y","-x+y)|^(2)],[(1.4)=|(1-lambda)x+lambda x+lambda y","(1-lambda)y-lambda x+lambda y|^(2)],[=|(x+lambda y","y-lambda x)|^(2)=(1+lambda^(2))x^(2)+(1+lambda^(2))y^(2)],[=(1+lambda^(2))|u|^(2) > |u|^(2).]:}\begin{align*} |(1-\lambda) u+\lambda T u|^{2} & =|(1-\lambda)(x, y)+\lambda T(x, y)|^{2} \\ & =|(1-\lambda)(x, y)+\lambda(x+y,-x+y)|^{2} \\ & =|(1-\lambda) x+\lambda x+\lambda y,(1-\lambda) y-\lambda x+\lambda y|^{2} \tag{1.4}\\ & =|(x+\lambda y, y-\lambda x)|^{2}=\left(1+\lambda^{2}\right) x^{2}+\left(1+\lambda^{2}\right) y^{2} \\ & =\left(1+\lambda^{2}\right)|u|^{2}>|u|^{2} . \end{align*}|(1−λ)u+λTu|2=|(1−λ)(x,y)+λT(x,y)|2=|(1−λ)(x,y)+λ(x+y,−x+y)|2(1.4)=|(1−λ)x+λx+λy,(1−λ)y−λx+λy|2=|(x+λy,y−λx)|2=(1+λ2)x2+(1+λ2)y2=(1+λ2)|u|2>|u|2.
Hence the Mann iteration is not convergent to ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0), for all { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1). Note that ( I 2 − t T ) − 1 I 2 − t T − 1 (I_(2)-tT)^(-1)\left(I_{2}-t T\right)^{-1}(I2−tT)−1 exists for all t ∈ ( 0 , 1 ) t ∈ ( 0 , 1 ) t in(0,1)t \in(0,1)t∈(0,1). Moreover,
(1.5) ( I 2 − t T ) = ( 1 − t − t t 1 − t ) ( I 2 − t T ) − 1 = 1 ( 1 − t ) 2 + t 2 ( 1 − t t − t 1 − t ) (1.5) I 2 − t T = 1 − t − t t 1 − t I 2 − t T − 1 = 1 ( 1 − t ) 2 + t 2 1 − t t − t 1 − t {:[(1.5)(I_(2)-tT)=([1-t,-t],[t,1-t])],[(I_(2)-tT)^(-1)=(1)/((1-t)^(2)+t^(2))([1-t,t],[-t,1-t])]:}\begin{gather*} \left(I_{2}-t T\right)=\left(\begin{array}{cc} 1-t & -t \\ t & 1-t \end{array}\right) \tag{1.5}\\ \left(I_{2}-t T\right)^{-1}=\frac{1}{(1-t)^{2}+t^{2}}\left(\begin{array}{cc} 1-t & t \\ -t & 1-t \end{array}\right) \end{gather*}(1.5)(I2−tT)=(1−t−tt1−t)(I2−tT)−1=1(1−t)2+t2(1−tt−t1−t)
Thus, for { α n } ∈ ( 0 , 1 ) α n ∈ ( 0 , 1 ) {alpha_(n)}in(0,1)\left\{\alpha_{n}\right\} \in(0,1){αn}∈(0,1),
(1.6) x n + 1 = 1 − α n ( 1 − α n ) 2 + α n 2 ( 1 − α n α n − α n 1 − α n ) x n (1.6) x n + 1 = 1 − α n 1 − α n 2 + α n 2 1 − α n α n − α n 1 − α n x n {:(1.6)x_(n+1)=(1-alpha_(n))/((1-alpha_(n))^(2)+alpha_(n)^(2))([1-alpha_(n),alpha_(n)],[-alpha_(n),1-alpha_(n)])x_(n):}x_{n+1}=\frac{1-\alpha_{n}}{\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}}\left(\begin{array}{cc} 1-\alpha_{n} & \alpha_{n} \tag{1.6}\\ -\alpha_{n} & 1-\alpha_{n} \end{array}\right) x_{n}(1.6)xn+1=1−αn(1−αn)2+αn2(1−αnαn−αn1−αn)xn
Denote x n := ( a n , b n ) ′ x n := a n , b n ′ x_(n):=(a_(n),b_(n))^(')x_{n}:=\left(a_{n}, b_{n}\right)^{\prime}xn:=(an,bn)′ to obtain
| x n + 1 | 2 = a n + 1 2 + b n + 1 2 = ( 1 − α n ) 2 [ ( 1 − α n ) 2 + α n 2 ] 2 [ ( ( 1 − α n ) a n + α n b n ) 2 + ( − α n a n + ( 1 − α n ) b n ) 2 ] = ( 1 − α n ) 2 [ ( 1 − α n ) 2 + α n 2 ] 2 [ ( 1 − α n ) 2 + α n 2 ] ( a n 2 + b n 2 ) = ( 1 − α n ) 2 [ ( 1 − α n ) 2 + α n 2 ] ( a n 2 + b n 2 ) (1.7) = ( 1 − α n 2 ( 1 − α n ) 2 + α n 2 ) | x n | 2 ≤ ( 1 − α n ) | x n | 2 x n + 1 2 = a n + 1 2 + b n + 1 2 = 1 − α n 2 1 − α n 2 + α n 2 2 1 − α n a n + α n b n 2 + − α n a n + 1 − α n b n 2 = 1 − α n 2 1 − α n 2 + α n 2 2 1 − α n 2 + α n 2 a n 2 + b n 2 = 1 − α n 2 1 − α n 2 + α n 2 a n 2 + b n 2 (1.7) = 1 − α n 2 1 − α n 2 + α n 2 x n 2 ≤ 1 − α n x n 2 {:[|x_(n+1)|^(2)=a_(n+1)^(2)+b_(n+1)^(2)],[=((1-alpha_(n))^(2))/([(1-alpha_(n))^(2)+alpha_(n)^(2)]^(2))[((1-alpha_(n))a_(n)+alpha_(n)b_(n))^(2)+(-alpha_(n)a_(n)+(1-alpha_(n))b_(n))^(2)]],[=((1-alpha_(n))^(2))/([(1-alpha_(n))^(2)+alpha_(n)^(2)]^(2))[(1-alpha_(n))^(2)+alpha_(n)^(2)](a_(n)^(2)+b_(n)^(2))=((1-alpha_(n))^(2))/([(1-alpha_(n))^(2)+alpha_(n)^(2)])(a_(n)^(2)+b_(n)^(2))],[(1.7)=(1-(alpha_(n)^(2))/((1-alpha_(n))^(2)+alpha_(n)^(2)))|x_(n)|^(2) <= (1-alpha_(n))|x_(n)|^(2)]:}\begin{align*} \left|x_{n+1}\right|^{2} & =a_{n+1}^{2}+b_{n+1}^{2} \\ & =\frac{\left(1-\alpha_{n}\right)^{2}}{\left[\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}\right]^{2}}\left[\left(\left(1-\alpha_{n}\right) a_{n}+\alpha_{n} b_{n}\right)^{2}+\left(-\alpha_{n} a_{n}+\left(1-\alpha_{n}\right) b_{n}\right)^{2}\right] \\ & =\frac{\left(1-\alpha_{n}\right)^{2}}{\left[\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}\right]^{2}}\left[\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}\right]\left(a_{n}^{2}+b_{n}^{2}\right)=\frac{\left(1-\alpha_{n}\right)^{2}}{\left[\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}\right]}\left(a_{n}^{2}+b_{n}^{2}\right) \\ & =\left(1-\frac{\alpha_{n}^{2}}{\left(1-\alpha_{n}\right)^{2}+\alpha_{n}^{2}}\right)\left|x_{n}\right|^{2} \leq\left(1-\alpha_{n}\right)\left|x_{n}\right|^{2} \tag{1.7} \end{align*}|xn+1|2=an+12+bn+12=(1−αn)2[(1−αn)2+αn2]2[((1−αn)an+αnbn)2+(−αnan+(1−αn)bn)2]=(1−αn)2[(1−αn)2+αn2]2[(1−αn)2+αn2](an2+bn2)=(1−αn)2[(1−αn)2+αn2](an2+bn2)(1.7)=(1−αn2(1−αn)2+αn2)|xn|2≤(1−αn)|xn|2
The last inequality holds because 0 ≥ 2 α n 2 − 3 α n + 1 0 ≥ 2 α n 2 − 3 α n + 1 0 >= 2alpha_(n)^(2)-3alpha_(n)+10 \geq 2 \alpha_{n}^{2}-3 \alpha_{n}+10≥2αn2−3αn+1 for α n ∈ ( 1 / 2 , 1 ) α n ∈ ( 1 / 2 , 1 ) alpha_(n)in(1//2,1)\alpha_{n} \in(1 / 2,1)αn∈(1/2,1). Lemma 1.1 assures that lim n → ∞ | x n | = 0 lim n → ∞   x n = 0 lim_(n rarr oo)|x_(n)|=0\lim _{n \rightarrow \infty}\left|x_{n}\right|=0limn→∞|xn|=0.
Take T : [ 0 , 1 ) → [ 0 , 1 ) , T x = x 2 T : [ 0 , 1 ) → [ 0 , 1 ) , T x = x 2 T:[0,1)rarr[0,1),Tx=x^(2)T:[0,1) \rightarrow[0,1), T x=x^{2}T:[0,1)→[0,1),Tx=x2, to obtain a map for which Mann iteration converges to the fixed point, while implicit Mann iteration is not well defined and consequently does not converge at all. Using (1.2),
(1.8) x n + 1 = ( 1 − α n ) x n + α n x n + 1 2 (1.8) x n + 1 = 1 − α n x n + α n x n + 1 2 {:(1.8)x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)x_(n+1)^(2):}\begin{equation*} x_{n+1}=\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} x_{n+1}^{2} \tag{1.8} \end{equation*}(1.8)xn+1=(1−αn)xn+αnxn+12
Solving for x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1 yields
(1.9) x n + 1 = 1 − ( 1 − 4 α n ( 1 − α n ) x n ) 1 / 2 2 α n (1.9) x n + 1 = 1 − 1 − 4 α n 1 − α n x n 1 / 2 2 α n {:(1.9)x_(n+1)=(1-(1-4alpha_(n)(1-alpha_(n))x_(n))^(1//2))/(2alpha_(n)):}\begin{equation*} x_{n+1}=\frac{1-\left(1-4 \alpha_{n}\left(1-\alpha_{n}\right) x_{n}\right)^{1 / 2}}{2 \alpha_{n}} \tag{1.9} \end{equation*}(1.9)xn+1=1−(1−4αn(1−αn)xn)1/22αn
or
(1.10) x n + 1 = 1 + ( 1 − 4 α n ( 1 − α n ) x n ) 1 / 2 2 α n (1.10) x n + 1 = 1 + 1 − 4 α n 1 − α n x n 1 / 2 2 α n {:(1.10)x_(n+1)=(1+(1-4alpha_(n)(1-alpha_(n))x_(n))^(1//2))/(2alpha_(n)):}\begin{equation*} x_{n+1}=\frac{1+\left(1-4 \alpha_{n}\left(1-\alpha_{n}\right) x_{n}\right)^{1 / 2}}{2 \alpha_{n}} \tag{1.10} \end{equation*}(1.10)xn+1=1+(1−4αn(1−αn)xn)1/22αn
In the later case, x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1 is no longer inside the interval [ 0 , 1 ) [ 0 , 1 ) [0,1)[0,1)[0,1). Suppose one always takes the first case. With the choice that each α n = 1 / 2 α n = 1 / 2 alpha_(n)=1//2\alpha_{n}=1 / 2αn=1/2, we have
(1.11) x n + 1 = 1 − ( 1 − x n ) 1 / 2 (1.11) x n + 1 = 1 − 1 − x n 1 / 2 {:(1.11)x_(n+1)=1-(1-x_(n))^(1//2):}\begin{equation*} x_{n+1}=1-\left(1-x_{n}\right)^{1 / 2} \tag{1.11} \end{equation*}(1.11)xn+1=1−(1−xn)1/2
that is, 1 − x n + 1 = ( 1 − x n ) 1 / 2 1 − x n + 1 = 1 − x n 1 / 2 1-x_(n+1)=(1-x_(n))^(1//2)1-x_{n+1}=\left(1-x_{n}\right)^{1 / 2}1−xn+1=(1−xn)1/2. Set a n = 1 − x n > 0 a n = 1 − x n > 0 a_(n)=1-x_(n) > 0a_{n}=1-x_{n}>0an=1−xn>0 to obtain lim n → ∞ a n = 0 lim n → ∞   a n = 0 lim_(n rarr oo)a_(n)=0\lim _{n \rightarrow \infty} a_{n}=0limn→∞an=0. Thus, { x n } x n {x_(n)}\left\{x_{n}\right\}{xn} converges to 1 , which is not in the interval [ 0 , 1 ) [ 0 , 1 ) [0,1)[0,1)[0,1).
The map J : X → 2 X ∗ J : X → 2 X ∗ J:X rarr2^(X^(**))J: X \rightarrow 2^{X^{*}}J:X→2X∗, given by J ( x ) := { f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ } J ( x ) := f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ J(x):={f inX^(**):(:x,f:)=||x||^(2),||f||=||x||}J(x):=\left\{f \in X^{*}:\langle x, f\rangle=\|x\|^{2},\|f\|=\|x\|\right\}J(x):={f∈X∗:⟨x,f⟩=‖x‖2,‖f‖=‖x‖}, for all x ∈ X x ∈ X x in Xx \in Xx∈X, is called the normalized duality mapping. It is easy to see that
(1.12) ⟨ y , j ( x ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) . (1.12) ⟨ y , j ( x ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) . {:(1.12)(:y","j(x):) <= ||x||||y||","quad AA x","y in X","AA j(x)in J(x).:}\begin{equation*} \langle y, j(x)\rangle \leq\|x\|\|y\|, \quad \forall x, y \in X, \forall j(x) \in J(x) . \tag{1.12} \end{equation*}(1.12)⟨y,j(x)⟩≤‖x‖‖y‖,∀x,y∈X,∀j(x)∈J(x).
Define
(1.13) Ψ := { ψ ∣ ψ : [ 0 , + ∞ ) ⟶ [ 0 , + ∞ ) is a strictly increasing map such that ψ ( 0 ) = 0 } . (1.13) Ψ := { ψ ∣ ψ : [ 0 , + ∞ ) ⟶ [ 0 , + ∞ )  is a strictly increasing map such that  ψ ( 0 ) = 0 } . {:(1.13)Psi:={psi∣psi:[0","+oo)longrightarrow[0","+oo)" is a strictly increasing map such that "psi(0)=0}.:}\begin{equation*} \Psi:=\{\psi \mid \psi:[0,+\infty) \longrightarrow[0,+\infty) \text { is a strictly increasing map such that } \psi(0)=0\} . \tag{1.13} \end{equation*}(1.13)Ψ:={ψ∣ψ:[0,+∞)⟶[0,+∞) is a strictly increasing map such that ψ(0)=0}.
The following definition can be found, for example, in [4].
Definition 1.3. A map T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X is called uniformly pseudocontractive if there exist a map ψ ∈ Ψ ψ ∈ Ψ psi in Psi\psi \in \Psiψ∈Ψ and a j ( x − y ) ∈ J ( x − y ) j ( x − y ) ∈ J ( x − y ) j(x-y)in J(x-y)j(x-y) \in J(x-y)j(x−y)∈J(x−y) such that
(1.14) ⟨ T x − T y , j ( x − y ) ⟩ ≤ ‖ x − y ‖ 2 − ψ ( ‖ x − y ‖ ) , ∀ x , y ∈ X . (1.14) ⟨ T x − T y , j ( x − y ) ⟩ ≤ ‖ x − y ‖ 2 − ψ ( ‖ x − y ‖ ) , ∀ x , y ∈ X . {:(1.14)(:Tx-Ty","j(x-y):) <= ||x-y||^(2)-psi(||x-y||)","quad AA x","y in X.:}\begin{equation*} \langle T x-T y, j(x-y)\rangle \leq\|x-y\|^{2}-\psi(\|x-y\|), \quad \forall x, y \in X . \tag{1.14} \end{equation*}(1.14)⟨Tx−Ty,j(x−y)⟩≤‖x−y‖2−ψ(‖x−y‖),∀x,y∈X.
Taking ψ ( a ) := ψ ( a ) ⋅ a ψ ( a ) := ψ ( a ) ⋅ a psi(a):=psi(a)*a\psi(a):=\psi(a) \cdot aψ(a):=ψ(a)⋅a, for all a ∈ [ 0 , + ∞ ) , ( ψ ∈ Ψ ) a ∈ [ 0 , + ∞ ) , ( ψ ∈ Ψ ) a in[0,+oo),(psi in Psi)a \in[0,+\infty),(\psi \in \Psi)a∈[0,+∞),(ψ∈Ψ), gives the usual definitions of ψ ψ psi\psiψ-strongly pseudocontractivity. The choice ψ ( a ) := y ⋅ a 2 , y ∈ ( 0 , 1 ) ψ ( a ) := y ⋅ a 2 , y ∈ ( 0 , 1 ) psi(a):=y*a^(2),y in(0,1)\psi(a):=y \cdot a^{2}, y \in(0,1)ψ(a):=y⋅a2,y∈(0,1), for all a ∈ [ 0 , + ∞ ) a ∈ [ 0 , + ∞ ) a in[0,+oo)a \in[0,+\infty)a∈[0,+∞), ( ψ ∈ Ψ ) ( ψ ∈ Ψ ) (psi in Psi)(\psi \in \Psi)(ψ∈Ψ), yields the usual definition of strong pseudocontractivity.
The convergence of (1.2) dealing with strongly pseudocontractive maps was proved in [8]. We will prove the convergence of iteration (1.2) when applied to uniformly pseudocontractive maps. For this purpose, we need the following result.
Lemma 1.4 [9]. Let { a n } a n {a_(n)}\left\{a_{n}\right\}{an} be a nonnegative bounded sequence which satisfies the following inequality:
(1.15) a n + 1 ≤ ( 1 − α n ) a n + α n a n + 1 − α n ψ ( a n + 1 ) a n + 1 + α n ε n , ∀ n ≥ n 0 (1.15) a n + 1 ≤ 1 − α n a n + α n a n + 1 − α n ψ a n + 1 a n + 1 + α n ε n , ∀ n ≥ n 0 {:(1.15)a_(n+1) <= (1-alpha_(n))a_(n)+alpha_(n)a_(n+1)-alpha_(n)(psi(a_(n+1)))/(a_(n+1))+alpha_(n)epsi_(n)","quad AA n >= n_(0):}\begin{equation*} a_{n+1} \leq\left(1-\alpha_{n}\right) a_{n}+\alpha_{n} a_{n+1}-\alpha_{n} \frac{\psi\left(a_{n+1}\right)}{a_{n+1}}+\alpha_{n} \varepsilon_{n}, \quad \forall n \geq n_{0} \tag{1.15} \end{equation*}(1.15)an+1≤(1−αn)an+αnan+1−αnψ(an+1)an+1+αnεn,∀n≥n0
where ψ ( ⋅ ) ∈ Ψ , α n ∈ ( 0 , 1 ) , ε n ≥ 0 ψ ( ⋅ ) ∈ Ψ , α n ∈ ( 0 , 1 ) , ε n ≥ 0 psi(*)in Psi,alpha_(n)in(0,1),epsi_(n) >= 0\psi(\cdot) \in \Psi, \alpha_{n} \in(0,1), \varepsilon_{n} \geq 0ψ(⋅)∈Ψ,αn∈(0,1),εn≥0, for all n ∈ N , ∑ n = 0 ∞ α n = ∞ n ∈ N , ∑ n = 0 ∞   α n = ∞ n inN,sum_(n=0)^(oo)alpha_(n)=oon \in \mathbb{N}, \sum_{n=0}^{\infty} \alpha_{n}=\inftyn∈N,∑n=0∞αn=∞, and lim n → ∞ ε n = 0 lim n → ∞   ε n = 0 lim_(n rarr oo)epsi_(n)=0\lim _{n \rightarrow \infty} \varepsilon_{n}=0limn→∞εn=0. Then lim n → ∞ a n = 0 lim n → ∞   a n = 0 lim_(n rarr oo)a_(n)=0\lim _{n \rightarrow \infty} a_{n}=0limn→∞an=0.

2. Main results

Theorem 2.1. Let X X XXX be a real Banach space and T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X a uniformly pseudocontractive map with a fixed point such that
(2.1) ∃ ( I − t T ) − 1 ∀ t ∈ ( 0 , 1 ) (2.1) ∃ ( I − t T ) − 1 ∀ t ∈ ( 0 , 1 ) {:(2.1)EE(I-tT)^(-1)quad AA t in(0","1):}\begin{equation*} \exists(I-t T)^{-1} \quad \forall t \in(0,1) \tag{2.1} \end{equation*}(2.1)∃(I−tT)−1∀t∈(0,1)
If { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) satisfies
(2.2) lim n → ∞ α n = 0 , ∑ n = 0 ∞ α n = ∞ (2.2) lim n → ∞   α n = 0 , ∑ n = 0 ∞   α n = ∞ {:(2.2)lim_(n rarr oo)alpha_(n)=0","quadsum_(n=0)^(oo)alpha_(n)=oo:}\begin{equation*} \lim _{n \rightarrow \infty} \alpha_{n}=0, \quad \sum_{n=0}^{\infty} \alpha_{n}=\infty \tag{2.2} \end{equation*}(2.2)limn→∞αn=0,∑n=0∞αn=∞
and { x n } x n {x_(n)}\left\{x_{n}\right\}{xn} is bounded, then for x 0 ∈ X x 0 ∈ X x_(0)in Xx_{0} \in Xx0∈X the iteration (1.2) converges to the fixed point of T T TTT.
Proof. The uniqueness of the fixed point comes from (1.14). Let x ∗ x ∗ x^(**)x^{*}x∗ be the fixed point of T T TTT.
If there exists a nonnegative integer n n nnn for which x n = x ∗ x n = x ∗ x_(n)=x^(**)x_{n}=x^{*}xn=x∗, then from (1.2),
(2.3) x n + 1 = ( 1 − α n ) x ∗ + α n T x n + 1 (2.3) x n + 1 = 1 − α n x ∗ + α n T x n + 1 {:(2.3)x_(n+1)=(1-alpha_(n))x^(**)+alpha_(n)Tx_(n+1):}\begin{equation*} x_{n+1}=\left(1-\alpha_{n}\right) x^{*}+\alpha_{n} T x_{n+1} \tag{2.3} \end{equation*}(2.3)xn+1=(1−αn)x∗+αnTxn+1
or
(2.4) ( I − α n T ) x n + 1 = ( I − α n T ) x ∗ (2.4) I − α n T x n + 1 = I − α n T x ∗ {:(2.4)(I-alpha_(n)T)x_(n+1)=(I-alpha_(n)T)x^(**):}\begin{equation*} \left(I-\alpha_{n} T\right) x_{n+1}=\left(I-\alpha_{n} T\right) x^{*} \tag{2.4} \end{equation*}(2.4)(I−αnT)xn+1=(I−αnT)x∗
which, using (2.1), implies that x n + 1 = x ∗ x n + 1 = x ∗ x_(n+1)=x^(**)x_{n+1}=x^{*}xn+1=x∗. By induction, x m = x ∗ x m = x ∗ x_(m)=x^(**)x_{m}=x^{*}xm=x∗ for all m ≥ n m ≥ n m >= nm \geq nm≥n.
We may therefore assume that each x n ≠ x ∗ x n ≠ x ∗ x_(n)!=x^(**)x_{n} \neq x^{*}xn≠x∗. Using (1.2)-(1.14),
‖ x n + 1 − x ∗ ‖ 2 = ⟨ x n + 1 − x ∗ , j ( x n + 1 − x ∗ ) ⟩ = ⟨ ( 1 − α n ) ( x n − x ∗ ) + α n ( T x n + 1 − T x ∗ ) , j ( x n + 1 − x ∗ ) ⟩ = ( 1 − α n ) ⟨ ( x n − x ∗ ) , j ( x n + 1 − x ∗ ) ⟩ + α n ⟨ T x n + 1 − T x ∗ , j ( x n + 1 − x ∗ ) ⟩ ≤ ( 1 − α n ) ‖ x n − x ∗ ‖ ‖ x n + 1 − x ∗ ‖ + α n ‖ x n + 1 − x ∗ ‖ 2 − α n ψ ( ‖ x n + 1 − x ∗ ‖ ) (2.5) ≤ ‖ x n + 1 − x ∗ ‖ ( ( 1 − α n ) ‖ x n − x ∗ ‖ + α n ‖ x n + 1 − x ∗ ‖ − α n ψ ( ‖ x n + 1 − x ∗ ‖ ) ‖ x n + 1 − x ∗ ‖ ) x n + 1 − x ∗ 2 = x n + 1 − x ∗ , j x n + 1 − x ∗ = 1 − α n x n − x ∗ + α n T x n + 1 − T x ∗ , j x n + 1 − x ∗ = 1 − α n x n − x ∗ , j x n + 1 − x ∗ + α n T x n + 1 − T x ∗ , j x n + 1 − x ∗ ≤ 1 − α n x n − x ∗ x n + 1 − x ∗ + α n x n + 1 − x ∗ 2 − α n ψ x n + 1 − x ∗ (2.5) ≤ x n + 1 − x ∗ 1 − α n x n − x ∗ + α n x n + 1 − x ∗ − α n ψ x n + 1 − x ∗ x n + 1 − x ∗ {:[||x_(n+1)-x^(**)||^(2)=(:x_(n+1)-x^(**),j(x_(n+1)-x^(**)):)],[=(:(1-alpha_(n))(x_(n)-x^(**))+alpha_(n)(Tx_(n+1)-Tx^(**)),j(x_(n+1)-x^(**)):)],[=(1-alpha_(n))(:(x_(n)-x^(**)),j(x_(n+1)-x^(**)):)+alpha_(n)(:Tx_(n+1)-Tx^(**),j(x_(n+1)-x^(**)):)],[ <= (1-alpha_(n))||x_(n)-x^(**)||||x_(n+1)-x^(**)||+alpha_(n)||x_(n+1)-x^(**)||^(2)-alpha_(n)psi(||x_(n+1)-x^(**)||)],[(2.5) <= ||x_(n+1)-x^(**)||((1-alpha_(n))||x_(n)-x^(**)||+alpha_(n)||x_(n+1)-x^(**)||-alpha_(n)(psi(||x_(n+1)-x^(**)||))/(||x_(n+1)-x^(**)||))]:}\begin{align*} \left\|x_{n+1}-x^{*}\right\|^{2} & =\left\langle x_{n+1}-x^{*}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ & =\left\langle\left(1-\alpha_{n}\right)\left(x_{n}-x^{*}\right)+\alpha_{n}\left(T x_{n+1}-T x^{*}\right), j\left(x_{n+1}-x^{*}\right)\right\rangle \\ & =\left(1-\alpha_{n}\right)\left\langle\left(x_{n}-x^{*}\right), j\left(x_{n+1}-x^{*}\right)\right\rangle+\alpha_{n}\left\langle T x_{n+1}-T x^{*}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ & \leq\left(1-\alpha_{n}\right)\left\|x_{n}-x^{*}\right\|\left\|x_{n+1}-x^{*}\right\|+\alpha_{n}\left\|x_{n+1}-x^{*}\right\|^{2}-\alpha_{n} \psi\left(\left\|x_{n+1}-x^{*}\right\|\right) \\ & \leq\left\|x_{n+1}-x^{*}\right\|\left(\left(1-\alpha_{n}\right)\left\|x_{n}-x^{*}\right\|+\alpha_{n}\left\|x_{n+1}-x^{*}\right\|-\alpha_{n} \frac{\psi\left(\left\|x_{n+1}-x^{*}\right\|\right)}{\left\|x_{n+1}-x^{*}\right\|}\right) \tag{2.5} \end{align*}‖xn+1−x∗‖2=⟨xn+1−x∗,j(xn+1−x∗)⟩=⟨(1−αn)(xn−x∗)+αn(Txn+1−Tx∗),j(xn+1−x∗)⟩=(1−αn)⟨(xn−x∗),j(xn+1−x∗)⟩+αn⟨Txn+1−Tx∗,j(xn+1−x∗)⟩≤(1−αn)‖xn−x∗‖‖xn+1−x∗‖+αn‖xn+1−x∗‖2−αnψ(‖xn+1−x∗‖)(2.5)≤‖xn+1−x∗‖((1−αn)‖xn−x∗‖+αn‖xn+1−x∗‖−αnψ(‖xn+1−x∗‖)‖xn+1−x∗‖)
Dividing by ‖ x n + 1 − x ∗ ‖ x n + 1 − x ∗ ||x_(n+1)-x^(**)||\left\|x_{n+1}-x^{*}\right\|‖xn+1−x∗‖ and defining a n = ‖ x n − x ∗ ‖ a n = x n − x ∗ a_(n)=||x_(n)-x^(**)||a_{n}=\left\|x_{n}-x^{*}\right\|an=‖xn−x∗‖ yield (1.15) with each ε n = 0 ε n = 0 epsi_(n)=0\varepsilon_{n}=0εn=0. From Lemma 1.4, lim n → ∞ a n = 0 lim n → ∞   a n = 0 lim_(n rarr oo)a_(n)=0\lim _{n \rightarrow \infty} a_{n}=0limn→∞an=0.
The following remark indicates some ways in which Theorem 2.1 can be applied to certain accretive maps.
Remark 2.2. (1) The operator T T TTT is a (uniformly, ψ ψ psi\psiψ-strongly, strongly) pseudocontractive map if and only if ( I − T I − T I-TI-TI−T ) is a (uniformly, ψ ψ psi\psiψ-strongly, strongly) accretive map.
(2) Let T , S : X → X T , S : X → X T,S:X rarr XT, S: X \rightarrow XT,S:X→X, and f ∈ X f ∈ X f in Xf \in Xf∈X be given. A fixed point for the map T x = f + ( I − S ) x T x = f + ( I − S ) x Tx=f+(I-S)xT x=f+(I-S) xTx=f+(I−S)x, for all x ∈ X x ∈ X x in Xx \in Xx∈X, is a solution for S x = f S x = f Sx=fS x=fSx=f, and conversely.
(3) Consider iteration (1.2) with T x = f + ( I − S ) x T x = f + ( I − S ) x Tx=f+(I-S)xT x=f+(I-S) xTx=f+(I−S)x to obtain a convergence result to the solution of S x = f S x = f Sx=fS x=fSx=f.
(4) Let f ∈ X f ∈ X f in Xf \in Xf∈X be given. If the operator S S SSS is accretive, then f − S f − S f-Sf-Sf−S is a strongly pseudocontractive map.
(5) Let T , S : X → X T , S : X → X T,S:X rarr XT, S: X \rightarrow XT,S:X→X. A fixed point for the map T x = f − S x T x = f − S x Tx=f-SxT x=f-S xTx=f−Sx, for all x ∈ X x ∈ X x in Xx \in Xx∈X, is a solution for x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f, and conversely.
(6) Consider iteration (1.2) with T x = f − S x T x = f − S x Tx=f-SxT x=f-S xTx=f−Sx to obtain a convergence result to the solution of x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f.
Remark 2.3. If (1.14) is also true for all x ∈ X x ∈ X x in Xx \in Xx∈X, and y := x ∗ y := x ∗ y:=x^(**)y:=x^{*}y:=x∗, the fixed point, then such a map is called uniformly hemicontractive. Obviously, our result holds for the uniformly hemicontractive case.

3. Remarks about implicit mean value iterations

Let X X XXX be a real Banach space and B B BBB a nonempty convex subset, u 0 , x 0 ∈ B u 0 , x 0 ∈ B u_(0),x_(0)in Bu_{0}, x_{0} \in Bu0,x0∈B. Consider for { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1), a finite family of functions { T i } i = 1 N : B → B T i i = 1 N : B → B {T_(i)}_(i=1)^(N):B rarr B\left\{T_{i}\right\}_{i=1}^{N}: B \rightarrow B{Ti}i=1N:B→B and the following two iterations:
x 1 = α 1 x 0 + ( 1 − α 1 ) T 1 x 1 , x 2 = α 2 x 1 + ( 1 − α 2 ) T 2 x 2 , ⋮ (3.1) x N = α N x N − 1 + ( 1 − α N ) T N x N , x N + 1 = α N + 1 x N + ( 1 − α N + 1 ) T 1 x N + 1 , ⋮ u 1 = α 1 u 0 + ( 1 − α 1 ) T 1 u 1 , u 2 = α 2 u 1 + ( 1 − α 2 ) T 2 u 2 , ⋮ x N = α N x N − 1 + ( 1 − α N ) T N x N , (3.2) x N + 1 = α N + 1 x N + ( 1 − α N + 1 ) T 1 2 x N + 1 , ⋮ x 2 N = α 2 N x 2 N − 1 + ( 1 − α 2 N ) T N 2 x 2 N , ⋮ x 1 = α 1 x 0 + 1 − α 1 T 1 x 1 , x 2 = α 2 x 1 + 1 − α 2 T 2 x 2 , ⋮ (3.1) x N = α N x N − 1 + 1 − α N T N x N , x N + 1 = α N + 1 x N + 1 − α N + 1 T 1 x N + 1 , ⋮ u 1 = α 1 u 0 + 1 − α 1 T 1 u 1 , u 2 = α 2 u 1 + 1 − α 2 T 2 u 2 , ⋮ x N = α N x N − 1 + 1 − α N T N x N , (3.2) x N + 1 = α N + 1 x N + 1 − α N + 1 T 1 2 x N + 1 , ⋮ x 2 N = α 2 N x 2 N − 1 + 1 − α 2 N T N 2 x 2 N , ⋮ {:[x_(1)=alpha_(1)x_(0)+(1-alpha_(1))T_(1)x_(1)","],[x_(2)=alpha_(2)x_(1)+(1-alpha_(2))T_(2)x_(2)","],[vdots],[(3.1)x_(N)=alpha_(N)x_(N-1)+(1-alpha_(N))T_(N)x_(N)","],[x_(N+1)=alpha_(N+1)x_(N)+(1-alpha_(N+1))T_(1)x_(N+1)","],[vdots],[u_(1)=alpha_(1)u_(0)+(1-alpha_(1))T_(1)u_(1)","],[u_(2)=alpha_(2)u_(1)+(1-alpha_(2))T_(2)u_(2)","],[vdots],[x_(N)=alpha_(N)x_(N-1)+(1-alpha_(N))T_(N)x_(N)","],[(3.2)x_(N+1)=alpha_(N+1)x_(N)+(1-alpha_(N+1))T_(1)^(2)x_(N+1)","],[vdots],[x_(2N)=alpha_(2N)x_(2N-1)+(1-alpha_(2N))T_(N)^(2)x_(2N)","],[vdots]:}\begin{gather*} x_{1}=\alpha_{1} x_{0}+\left(1-\alpha_{1}\right) T_{1} x_{1}, \\ x_{2}=\alpha_{2} x_{1}+\left(1-\alpha_{2}\right) T_{2} x_{2}, \\ \vdots \\ x_{N}=\alpha_{N} x_{N-1}+\left(1-\alpha_{N}\right) T_{N} x_{N}, \tag{3.1}\\ x_{N+1}=\alpha_{N+1} x_{N}+\left(1-\alpha_{N+1}\right) T_{1} x_{N+1}, \\ \vdots \\ u_{1}=\alpha_{1} u_{0}+\left(1-\alpha_{1}\right) T_{1} u_{1}, \\ u_{2}=\alpha_{2} u_{1}+\left(1-\alpha_{2}\right) T_{2} u_{2}, \\ \vdots \\ x_{N}=\alpha_{N} x_{N-1}+\left(1-\alpha_{N}\right) T_{N} x_{N}, \\ x_{N+1}=\alpha_{N+1} x_{N}+\left(1-\alpha_{N+1}\right) T_{1}^{2} x_{N+1}, \tag{3.2}\\ \vdots \\ x_{2 N}=\alpha_{2 N} x_{2 N-1}+\left(1-\alpha_{2 N}\right) T_{N}^{2} x_{2 N}, \\ \vdots \end{gather*}x1=α1x0+(1−α1)T1x1,x2=α2x1+(1−α2)T2x2,⋮(3.1)xN=αNxN−1+(1−αN)TNxN,xN+1=αN+1xN+(1−αN+1)T1xN+1,⋮u1=α1u0+(1−α1)T1u1,u2=α2u1+(1−α2)T2u2,⋮xN=αNxN−1+(1−αN)TNxN,(3.2)xN+1=αN+1xN+(1−αN+1)T12xN+1,⋮x2N=α2Nx2N−1+(1−α2N)TN2x2N,⋮
Iteration (3.1) has been considered in [2, 6, 7, 11, 12]. Iteration (3.1) has been discussed in [ 1 , 10 ] [ 1 , 10 ] [1,10][1,10][1,10]. Note that iteration (1.2) is a particular case of (3.1). However, as far as we know, no such paper is dedicated to the convergence of the implicit iteration dealing with uniformly pseudocontractive maps.
Condition (2.1) forces iteration (1.2) to be well defined. The papers listed above do not impose such a condition, and consequently, the resulting implicit mean value iterations need not be well defined, as the following example illustrates.
Example 3.1. Take T i : [ 0 , 1 ] → [ 0 , 1 ] , i = 1 , 2 , T 1 ( x ) = x 2 T i : [ 0 , 1 ] → [ 0 , 1 ] , i = 1 , 2 , T 1 ( x ) = x 2 T_(i):[0,1]rarr[0,1],i=1,2,T_(1)(x)=x^(2)T_{i}:[0,1] \rightarrow[0,1], i=1,2, T_{1}(x)=x^{2}Ti:[0,1]→[0,1],i=1,2,T1(x)=x2, and T 2 ( x ) = ( 1 / 2 ) x T 2 ( x ) = ( 1 / 2 ) x T_(2)(x)=(1//2)xT_{2}(x)=(1 / 2) xT2(x)=(1/2)x with the common fixed point x ∗ = 0 x ∗ = 0 x^(**)=0x^{*}=0x∗=0. Then for x 0 = 1 , x 1 = α 1 x 0 + ( 1 − α 1 ) x 1 2 x 0 = 1 , x 1 = α 1 x 0 + 1 − α 1 x 1 2 x_(0)=1,x_(1)=alpha_(1)x_(0)+(1-alpha_(1))x_(1)^(2)x_{0}=1, x_{1}=\alpha_{1} x_{0}+\left(1-\alpha_{1}\right) x_{1}^{2}x0=1,x1=α1x0+(1−α1)x12, one obtains x 1 = x 1 = x_(1)=x_{1}=x1= 1 and x 1 = α 1 / ( 1 − α 1 ) x 1 = α 1 / 1 − α 1 x_(1)=alpha_(1)//(1-alpha_(1))x_{1}=\alpha_{1} /\left(1-\alpha_{1}\right)x1=α1/(1−α1). Take now u 0 = 1 , u 1 = α 1 u 0 + ( 1 − α 1 ) u 1 2 u 0 = 1 , u 1 = α 1 u 0 + 1 − α 1 u 1 2 u_(0)=1,u_(1)=alpha_(1)u_(0)+(1-alpha_(1))u_(1)^(2)u_{0}=1, u_{1}=\alpha_{1} u_{0}+\left(1-\alpha_{1}\right) u_{1}^{2}u0=1,u1=α1u0+(1−α1)u12, and u 2 = α 2 u 1 + ( 1 − α 2 ) ( 1 / 4 ) u 2 u 2 = α 2 u 1 + ( 1 − α 2 ( 1 / 4 ) u 2 u_(2)=alpha_(2)u_(1)+(1-{:alpha_(2))(1//4)u_(2)u_{2}=\alpha_{2} u_{1}+(1- \left.\alpha_{2}\right)(1 / 4) u_{2}u2=α2u1+(1−α2)(1/4)u2. Observe that there are two possible values for u 2 u 2 u_(2)u_{2}u2.
Remark 3.2. The existence of ( I − t T i ) − 1 I − t T i − 1 (I-tT_(i))^(-1)\left(I-t T_{i}\right)^{-1}(I−tTi)−1, for all t ∈ ] 0 , 1 [ , i = 1 , N t ∈ 0 , 1 [ , i = 1 , N {:t in]0,1[,i=1,N\left.t \in\right] 0,1[, i=1, Nt∈]0,1[,i=1,N, should be added to the hypotheses of the results of [ 2 , 6 , 7 , 11 , 12 ] [ 2 , 6 , 7 , 11 , 12 ] [2,6,7,11,12][2,6,7,11,12][2,6,7,11,12] in order to have well-defined iterations. The existence of ( I − t T i p ) − 1 I − t T i p − 1 (I-tT_(i)^(p))^(-1)\left(I-t T_{i}^{p}\right)^{-1}(I−tTip)−1, for all t ∈ ( 0 , 1 ) , i = 1 , N t ∈ ( 0 , 1 ) , i = 1 , N t in(0,1),i=1,Nt \in(0,1), i=1, Nt∈(0,1),i=1,N, for all p ≥ 1 p ≥ 1 p >= 1p \geq 1p≥1, should be added to the hypotheses of the results of [1,10].

Acknowledgment

The authors are indebted to the referees for carefully reading the paper and for making useful suggestions.

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[10] Z.-H. Sun, Strong convergence of an implicit iteration process for a finite family of asymptotically quasi-nonexpansive mappings, Journal of Mathematical Analysis and Applications 286 (2003), no. 1, 351-358.
[11] H.-K. Xu and R. G. Ori, An implicit iteration process for nonexpansive mappings, Numerical Functional Analysis and Optimization 22 (2001), no. 5-6, 767-773.
[12] Y. Zhou and S. S. Chang, Convergence of implicit iteration process for a finite family of asymptotically nonexpansive mappings in Banach spaces, Numerical Functional Analysis and Optimization 23 (2002), no. 7-8, 911-921.
B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA E-mail address: rhoades@indiana.edu
Ştefan M. Şoltuz: "Numerical Analysis and Application Theory Workgroup, Tiberiu Popoviciu Institute of Numerical Analysis," P.O. Box 68-1, 400110 Cluj-Napoca, Romania
E-mail address: smsoltuz@gmail.com
2006

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